solution chemistry dealing with mixtures 1. solutions a solution is a homogenous mixture consisting...
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Solutions
A solution is a homogenous mixture consisting of a solvent and at least one solute.
The solvent is the most prevalent species.
The solute is the less prevalent species.
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Examples of Solutions
Saline (salt water) is a solution. The solvent is water, the solute is salt.
Wet salt is also a solution. The solvent is salt, the solute is water.
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160 proof Vodka
What is the solvent?
Alcohol – It is 80% alcohol.
What is the solute?
Water – It is 20% water.
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Aqueous Solutions
Aqueous solutions are specifically solutions where water is the solvent.
Aqueous solutions are a very common medium for performing chemical reactions.
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Advantages of Aqueous Solutions
1. Mixing – you can stir the solution.
2. Ability to dissipate heat (or cold) – the mass of the solvent allows it to absorb significant amounts of heat (or cold).
3. “Universal solvent” – water dissolves many different materials, especially ionic materials.
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Concentration
Because a solution is a mixture – there are different ratios of solvent/solute quantities possible.
For example, I could put 1 teaspoon of salt in a cup of water OR I could put 2 teaspoons of salt in a cup of water.
Both are saline solutions, but they have different amounts of salt.
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Concentration
Almost any unit of measure can be used to specify concentration. (teaspoon solute/cup solvent would work!)
There are certain common units of measuring solution concentration that are most frequently used. Understanding their UNITS! UNITS! UNITS! And being able to manipulate those UNITS! UNITS! UNITS! is crucial.
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Common units of concentration
Normality
ppt –
ppm –
ppb –
lb/million gallons -
% by mass –
% by volume
% by mass-volume
Molarity –
Molality –
Converting unitsWhat is the molarity of a 10% by mass aqueous NaCl solution?
UNITS! UNITS! UNITS!
10 g NaCl = moles NaCl100 g NaCl solution L solution
To convert g NaCl to moles, you need to know…Molar mass of NaCl
To convert g solution to L solution, you need to know…Density of the solution
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How much sugar can I put in a cup (8 oz) of boiling coffee?
A. 8 Oz
B. 4 teaspoons
C. 16 oz
D. 2 oz
E. It doesn’t matter – says the cynical student in the back row.
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The Density
You don’t always know the density.
Density depends on concentration.
Sometimes you know the density.
Sometimes you can figure out the density.
Sometimes you just have to ASSUME the density.
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The Density
If you don’t know anything except what was given:
What is the molarity of a 10% by mass aqueous NaCl solution?
What would you do?
Assume the density is that of pure water (1.0 g/mL at 25°C)
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The Density
Suppose I had further information:
What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL)
Now what would you do?
I can either ASSUME that 5% is “close enough” to 10%.OR I can “interpolate” the density between 5% and 20%.
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Linear Interpolation
Do you know what a “linear interpolation is”?
I assume that there is a linear (straight-line) dependence of the density on the concentration. (By the way, this is not true, but it is an OK assumption if the range is narrow enough.)
Then I draw a straight line between the two points I know and find the interpolated concentration at my concentration of interest.
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% by mass NaCl Density (g/mL) 5 1.05
20 1.13
y = 0.0053x + 1.0233
1.04
1.05
1.06
1.07
1.08
1.09
1.1
1.11
1.12
1.13
1.14
0 5 10 15 20 25
Density (g/mL)
Density (g/mL)
Linear (Density (g/mL) )
You can also do it without the graph.
What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL)
I find the slope of the line: Δ Density Δ % NaCl1.13 g/mL – 1.05 g/mL = 5.33x10-3 g/mL 20% - 5 % %
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My ProblemWhat is the molarity of a 10% by mass aqueous NaCl solution? (Density of
5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL)
5.33x10-3 g/mL means that every 1% change in concentration % results in a 5.33x10-3 g/mL change in density
10%-5% = 5% change5.33x10-3 g/mL * 5 % = 0.0267 g/mL change %
1.05 g/mL + 0.0267 g/mL = 1.077 g/mL = 1.08 g/mL interpolated density
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Solving the problem
What is the molarity of a 10% by mass aqueous NaCl solution? (Density of 5% NaCl solution = 1.05 g/mL, Density of 20% NaCl solution = 1.13 g/mL)
10 g NaCl * 1 mol NaCl = 0.171 mol NaCl100 g solution 58.45 g NaCl 100 g solution
0.171 mol NaCl * 1.08 g solution * 1000 mL = 1.84 mol NaCl100 g solution 1 mL solution 1 L L solution
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Converting units
Typically speaking, you can convert any of the concentration units into any of the others as long as you have the Molar Mass and the Density!
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Density – your critical judgment
For a solution, sometimes you know the density, sometimes you don’t.
There are tables, but they are not all inclusive.
You might, for example, find in a table that:Density (30% HCl) = 1.12 g/mLDensity (40% HCl) = 1.23 g/mLDensity (36% HCl) = ???
Interpolate or Assume
Density (30% HCl) = 1.12 g/mL
Density (40% HCl) = 1.23 g/mL
Density (36% HCl) = ???
You could assume that 36% is closest to 40% and use 1.23 g/mL. This is legitimate, although not 100% accurate. Results may vary, depending on how good the assumption is.
Interpolate or AssumeDensity (30% HCl) = 1.12 g/mL Density (40% HCl) = 1.23 g/mLDensity (36% HCl) = ???
You could assume that density changes linearly with concentration (it doesn’t, but it is pseudo-linear for small changes). In that case, you would “linearly interpolate” the density.
1.23 g/mL – 1.12 g/mL = 0.011 g/mL = 0.011 g 40% HCl-30%HCl % mL%
1.12 g/mL + 0.011 g/mL% * 6% = 1.186 g/mL = 1.19 g/mL
This is legitimate, although still not 100% accurate, but probably better than the previous assumption.
If I don’t have Density tables…
For dilute solutions, you can get pretty close by assuming the density of the solution is the same as the density of pure water.
For concentrated solutions (like 36%), this is probably not a good assumption, but it is better than nothing!
Further Example
56.0 g of Fe2O3 was dissolved in water yielding a total solution volume of 2.65 L. What is the molarity of the resulting solution?
56.0 g Fe2O3 * 1 mol Fe2O3 = 0.351 mol Fe2O3
159.69 g Fe2O3
0.351 mol Fe2O3 = 0.132 M Fe2O3
2.65 L
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An example
56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?
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Limiting Reagent Problem
What’s the first thing you need?
A balanced equation!
(NH4)2SO4 + FeCl3 → Fe2(SO4)3 + NH4Cl
How do you know this is the right products?
Charges! This is an example of a double replacement reaction. The cations get switched (or the anions, if you prefer).
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Limiting Reagent Problem
We still need to balance it!
(NH4)2SO4 + FeCl3 → Fe2(SO4)3 + NH4Cl
3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl
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Armed with Stoichiometry!
56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?
3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl
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Armed with Stoichiometry!
56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?
3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl
36.0 g FeCl3 *1 mol FeCl3 * 1 mol Fe2(SO4)3 * 399.87 g Fe2(SO4)3= 44.37 g Fe2(SO4)3
162.21 g FeCl3 2 mol FeCl3 1 mol Fe2(SO4)3
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Armed with Stoichiometry!
56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?
3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl
Like any ratio of units, this is really just a conversion factor!!!
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Armed with Stoichiometry!
56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?
3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl
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Armed with Stoichiometry!
56.50 mL of a 2.15 M ammonium sulfate solution is mixed with 36.0 g of iron (III) chloride. If the reaction proceeds with a 65% yield, how much iron (III) sulfate would be acquired?
3 (NH4)2SO4 + 2 FeCl3 → Fe2(SO4)3 + 6 NH4Cl
Limiting Reagent is (NH4)2SO4: 16.13 g Fe2(SO4)3 theoretical
16.13 g Fe2(SO4)3 theoretical * 65 g actual = 10.48 g actual Fe2(SO4)3
100 g theoretical
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Clicker Question
I have 1 L of a solution that is 5.4% by mass sodium sulfate. If the density of 5% sodium sulfate is 1.085 g/mL, how much silver (I) chloride would I need to add to precipitate all of the sulfate?
A. 59 g
B. 257 g
C.118 g
D. 129 g
E. 25.4 g
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Na2SO4 + 2 AgCl 2 NaCl + Ag2SO4
1L * 1000 mL * 1.085 g *5.4 g Na2SO4 = 58.59 g Na2SO4
1L 1 mL 100 g solution
58.59 g Na2SO4 * 1 mol Na2SO4 = 0.4126 mol Na2SO4
142 g Na2SO4
0.4126 mol Na2SO4 * 2 mol AgCl = 0.825 mol AgCl
1 mol Na2SO4
0.825 mol AgCl * 143 g AgCl = 118 g AgCl 1 mol AgCl
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Question
When 50.00 mL of 0.125 mol/L silver (I) nitrate is mixed with 50.00 mL of 0.250 M sodium sulfate a greyish solid forms. If I recover 0.813 g of solid, what is the yield of the reaction?
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